Biogechemical Cycles and Atmospheric Oxygen

COMPUTER LAB 4

EARTH SYSTEMS SCIENCE I

PG251, Spring 2011

Lab 4. Basic Concepts in Chemistry for Studying Biogeochemical Cycles

In this lab we learn some very basic concepts in chemistry that will allow you a much better understanding of the functioning of biogeochemical cycles and other topics discussed in this class. This will be completely review for those of you who have already taken chemistry, and it will be a very brief introduction for those who have not yet taken a chemistry course. Additional material, if you would like some more explanation, can be found at these fantastic web sites from the Visionlearning project:

Stoichiometry & Chemical Equations
http://www.visionlearning.com/library/module_viewer.php?mid=56&l=&c3=
The Mole and Molecular Weight
http://www.visionlearning.com/library/module_viewer.php?mid=53&l=&c3=

a. Stoichiometry. Chemical equations, or reactions, such as the one representing photosynthesis which we have already discussed (equation 1), are “balanced.” This means that the relative numbers of molecules involved in the reaction are given by the numerical coefficients preceding the chemical symbol for the molecule. So, in equation (1), 6 molecules of CO2 gas react with 6 molecules of H2O liquid to form one molecule of C6H12O6 (glucose) and six molecules of molecular oxygen (O2). Another way of saying this is that the stoichiometric ratio of the reactant CO2 to the product O2 is 6:6, or 1:1. It also means that the number of atoms of each element is the same in the reactants (on the left hand side of the equation) and in the products (on the right hand side of the equation). In equation 1, there are 18 oxygen atoms, 6 carbon atoms, and 12 hydrogen atoms in both the reactants and in the products (i.e. on each side of the equation). Thus, mass is conserved in stoichiometric equations.

6CO2 (g) + 6H2O (l) à C6H12O6 (s) + 6O2 (g) (1)

b. Atomic and molecular weights. Each element on the periodic table has an atomic weight assigned to it. These are unitless, or dimensionless, numbers that indicate how much mass an atom contains compared to carbon-12 (which has six protons and six neutrons). The molecular weight of a molecule is the sum of the atomic weights of all the atoms in the molecule. So, the molecular weight of water (H2O) is equal to 2 times the molecular weight of hydrogen (2x1.008) plus the molecular weight of oxygen (15.999) for a total of 18.015 (see table 1).

Table 1. Atomic weights of selected elements.

c. Gram-molecular weights, or moles or mols. One mole of any compound contains Avogadro’s number (NA) of molecules, where NA = 6.022e23, or 6.022 x 1023 molecules. It turns out that the mass, in grams, of one mole of any substance is equal to its molecular weight. For example, one mole of water has a mass of 18.015 grams. So, one can think of the atomic and molecular weights of having units grams/mole. This allows us to use equation 1, where we have a relationship between the number of molecules of each substance in a chemical reaction, to discussing the number of moles of each substance in a chemical reaction: 6 moles of CO2 gas react with 6 moles of H2O liquid to form one mole of C6H12O6 (glucose) and six moles of O2 (oxygen). If we multiply each coefficient by the molecular weight of the molecule, we can estimate how many grams of each molecule are involved in the reaction. Note that, because each molecule has a different molecular weight, the ratios of the masses of each molecule will be different than the ratios of the number of moles.

For example, if 100 moles of CO2 are taken up by a photosynthesizing plant, how many grams of glucose are created? According to the stoichiometry of equation 1, for every mole of CO2 there are (1/6) moles of glucose created. The molecular weight of glucose (C6H12O6) is 6x12.011 + 12x1.008 + 6x15.999 = 180.16 grams per mole. So, to calculate how many grams, or kilograms, of glucose, multiply the number of moles by the molecular weight as in equation 2:

(100/6) [moles of glucose] * 180.16 [grams/mole] = 3002.6 grams @ 3 kg (2)

For another example, in our lab last semester on the carbon cycle, we estimated the pre-industrial equilibrium value of CO2 in the atmosphere to be @ 600 Gt-C, or 600E15 grams of carbon. We will now answer three questions: how many moles of carbon is this? how many moles of carbon dioxide is this? how many grams of carbon dioxide is this? First, to convert units from grams to moles, one has to divide by the molecular weight (which has units grams/mole) as shown in equation 3:

#moles of Carbon @ 600E15 [g-C] / 12 [g/mol] @ 50E15 [moles] (3)

How many moles of carbon dioxide is this? Well, for each mole of C in CO2 there is exactly one mole of CO2. So, there are 50E15 moles of CO2 in the pre-industrial atmosphere. How many grams of CO2 is this? To answer that question we must first calculate the molecular weight of CO2, which is the sum of the atomic weights of each atom in the molecule: the atomic weight of carbon plus two times the atomic weight of oxygen: 12.011 + 2x15.999 @ 44 [g/mol]. To estimate how many grams of CO2 are in the atmosphere, multiply the number of moles by the molecular weight as in equation 4:

#grams of CO2 @ 50E15 [moles] * 44 [g/mol] @ 2200E15 [g CO2] (4)

EXERCISES: All exercises CAN be done in an EXCEL spreadsheet but it is not necessary – what is necessary is to CLEARLY show all calculations!

1.  How much gas is there in the atmosphere? Using the concepts discussed in this lab, and the values provided to you on table 2, estimate and fill in the missing values (yellow cells) for each gas on table 2. Note that concentrations are usually expressed by volume, which is the same as saying that “if 78% of the atmosphere is nitrogen, then 78% of the molecules in the air are nitrogen, and 78% of the moles in the air are nitrogen.” It does not necessarily mean that 78% of the weight of the gasses in the air is from nitrogen. Also note that these values do not include water vapor, and that the molecular weight of molecules that have two atoms of the same element (e.g. N2, O2) is twice the atomic weight. These same principals can be used in estimating the masses of elements in the lithosphere and hydrosphere as well, but that’s not part of this assignment. Hand in your answer so that I can see your calculations. If I can not see your calculations, I will not give you credit for the answer. Are the numbers that you get for carbon dioxide consistent with the value we used to initialize the atmospheric carbon stock in the carbon cycle lab last semester? (Note – in the carbon cycle lab, the units of the stocks were in grams of carbon, not carbon dioxide.)

Table 2. Information on selected constituents of dry air. Yellow cells are to be filled in by the student.

* the mean molecular weight of dry air is equal to the concentration-weighted mean of the molecular weights of all the constituents.

**concentration of carbon dioxide set at pre-industrial level.

2. Air pressure, kilograms, and pounds. This question is a bit of a digression into atmospheric physics, not chemistry, but is relevant nevertheless. It was inspired by a question asked by a student in this class a few years ago, so you can thank her for this opportunity or blame me if you find it annoying. In the previous question I gave you the approximate dry mass of the atmosphere. In this question, you will use that number to estimate the mean air pressure at the earth’s surface. You should be able to compare your estimate to what we know is the actual number (equation 5):

Mean surface air pressure on earth expressed in different units:

@ 1013 millibars (mb)

@ 1013 hecto-Pascals (hPa)

@ 1013 x 102 Pa

@ 1.013 x 105 Pa

@ 1.013 x 105 kg/(m s2) *** standard units ***

@ 1 atmosphere (atm) (5)

A number of different units are commonly used to describe air pressure. The only units that we can calculate directly here are the ones highlighted in bold, which tells you that one pascal is equal to one kg/(m s2). This unit may not make intuitive sense to you, but perhaps when you see how it is derived (in the next paragraph) it might at least make some sense. First, an important point is that the air pressure at the surface of the earth is equal to the weight of all the air above that point. In fact, the air pressure at any level in the atmosphere is equal to the weight of all the air above that point.

Pressure is a force. If you remember from way back in the beginning of last semester, Newton’s first law states that “F = ma”: force is equal to mass times acceleration. In standard units, mass is kg, and acceleration is m/s2. If you correct for the fact that we are calculating everything “per square meter”, and then do a unit check on that multiplication, you will find that the units are correct (equation 6):

[kg] * [m/s2] / [m2] = [k/(m s2)] (6)

On earth, when we are talking about air pressure and weight, the rate of acceleration is a well known number usually referred to as g, the acceleration due to gravity. Your assignment is to estimate the mean surface air pressure on earth. You are given the mass of the entire atmosphere, and the surface area of the earth, as well as the value of g. Use the number given in table 2 as an estimate of the dry mass of the atmosphere, and assume that the water in the atmosphere is equal to 4% of the total atmospheric mass (not 4% of the dry mass). Fill in the missing values (yellow cells) on table 3. Considering that the numbers we use for the total mass of dry air and water vapor are approximate, your estimate should be a good approximation to the real number shown in equation 5. Hand in your answer so that I can see your calculations. If I can not see your calculations, I will not give you credit for the answer.

Table 3. Calculations for estimating the average air pressure at the surface of the earth. Yellow cells are to be filled in by the student.

* mass of dry air should be taken from table 2

** assume that the mass of water vapor is approximately 4% of the total mass of the atmosphere (not 4% of the mass of dry air).

The difference between weight and mass: Now, here’s a funny thing about air pressure. There is something fundamentally different about air pressure when expressed in British units (i.e. pounds and inches) versus standard units. In British units the mean surface air pressure is usually estimated at @ 14.7 lbs/inch2. A unit of “weight” (lbs) is in the numerator, and a unit of area (inches2) is in the denominator. In standard units, when we have kg in the numerator and m2 in the denominator, we have to multiply by the acceleration of gravity to derive the air pressure. Why don’t we have to multiply by an acceleration term when using British units?

This is because lbs and kgs are not actually different units for the same physical property. The British unit pound is a weight, which includes the acceleration term already, while a kg is a mass, which does not include the acceleration term. For example, on the moon, where the gravitational acceleration is approximately 1/6 of the value on earth, your weight would be 1/6 of your weight on earth (and, if we had the same amount of air in the atmosphere, the air pressure would be 1/6 of that on earth). This means that the force your body would exert on the surface would be less on the moon than on earth. Your mass, however, which is a function of the number of protons and neutrons in your body, does not change whether you are on earth, on the moon, floating around in space, or anywhere else! This is the difference between mass and weight.

But in most countries the unit kg is used just as we use weights here. That’s because it is assumed that the acceleration of gravity is a constant value, so that either kg (a unit of mass) or lbs (a unit of weight) can be used.

3. Eutrophication and dissolved oxygen. This question is meant to serve as a brief introduction to the problem of eutrophication, and the way stoichiometry is sometimes used to estimate the effect of nutrients on biological activity and the subsequent effect on biogeochemical cycles, and in this case on dissolved oxygen in water.

In natural water bodies, such as lakes or coastal areas, as time passes, increasing levels of dissolved nutrients are often found in the water. These dissolved nutrients are washed in from the surrounding land surface, or sometimes deposited on the water body directly from the atmosphere. These nutrients encourage the growth of aquatic plants, sometimes called algal blooms. When the plants die, bacterial respiration (i.e. decomposition) occurs, which requires the removal of oxygen that is dissolved in the water. The rate of dissolved oxygen depletion by bacterial respiration is called biological oxygen demand (BOD). If there is no source of oxygen to replenish the BOD, then the concentrations of dissolved oxygen in the water body can decrease to levels that are too low to support aquatic life, a condition sometimes called hypoxia.